Let a function f:[0,5] rightarrow R be contin | vedclass

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(A) Given $F(x) = \int_{1}^{x} t^{2} g(t) dt$. By the Fundamental Theorem of Calculus,$F'(x) = x^{2} g(x) = x^{2} \int_{1}^{x} f(u) du$. At $x=1$,$F'(

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a point of local minima

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(A) Given $F(x) = \int_{1}^{x} t^{2} g(t) dt$. By the Fundamental Theorem of Calculus,$F'(x) = x^{2} g(x) = x^{2} \int_{1}^{x} f(u) du$. At $x=1$,$F'(1) = 1^{2} \int_{1}^{1} f(u) du = 0$. Thus,$x=1$ is a critical point. Now,find the second derivative: $F''(x) = \frac{d}{dx} [x^{2} g(x)] = x^{2} g'(x) + 2x g(x)$. Since $g(t) = \int_{1}^{t} f(u) du$,$g'(t) = f(t)$. So,$F''(x) = x^{2} f(x) + 2x \int_{1}^{x} f(u) du$. Evaluating at $x=1$: $F''(1) = 1^{2} f(1) + 2(1) \int_{1}^{1} f(u) du = f(1) + 0 = 3$. Since $F'(1) = 0$ and $F''(1) = 3 > 0$,by the Second Derivative Test,$x=1$ is a point of local minima.

Let a function $f:[0,5] \rightarrow R$ be continuous. $f(1)=3$ and $F$ be defined as $F(x)=\int_{1}^{x} t^{2} g(t) dt$,where $g(t)=\int_{1}^{t} f(u) du$. Then for the function $F$,the point $x=1$ is

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